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KEAM 2010 Math Paper

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Question : 6 of 120

Marks: +1, -0

If

\vec{p}, \vec{q}

\vec{r}

are perpendicular to

\vec{q}+\vec{r}, \vec{r}+\vec{p}

\vec{p}+\vec{q}

respectively and if

|\vec{p}+\vec{q}|=6

|\vec{q}+\vec{r}|=4\sqrt{3}

|\vec{r}+\vec{p}|=4, \;\;

|\vec{p}+\vec{q}+\vec{r}|

is

$5\sqrt{2}$
10
15
5
25

Solution:

\vec{p}\perp(\vec{q}+\vec{r}) \Rightarrow \vec{p}\cdot(\vec{q}+\vec{r})=0

\Rightarrow \vec{p}\cdot \vec{q}+\vec{p}\cdot \vec{r}=0

...(i)

\vec{q}\perp(\vec{r}+\vec{p}) \Rightarrow \vec{q}\cdot(\vec{r}+\vec{p})=0

\Rightarrow \vec{q}\cdot \vec{r}+\vec{q}\cdot \vec{p}=0

....(ii)

\vec{r}\perp(\vec{p}+\vec{q}) \Rightarrow \vec{r}\cdot(\vec{p}+\vec{q})=0

\Rightarrow \vec{r}\cdot \vec{p}+\vec{r}\cdot \vec{q}=0

....(iii)

Adding Eqs. (i), (ii) and (iii), we get

\vec{p}\cdot \vec{q}+\vec{q}\cdot \vec{r}+\vec{r}\cdot \vec{p}=0

.........(iv)

Now,

|\vec{p}+\vec{q}|=6 \Rightarrow (\vec{p}+\vec{q})\cdot(\vec{p}+\vec{q})=36

\Rightarrow |\vec{p}|^{2}+\vec{p}\cdot \vec{q}+\vec{q}\cdot \vec{p}+|\vec{q}|^{2}=36

........(v)

Similarly,

|\vec{q}+\vec{r}|=4\sqrt{3}

\Rightarrow |\vec{q}|^{2}+\vec{q}\cdot \vec{r}+\vec{r}\cdot \vec{q}+|\vec{r}|^{2}=48

.....(vi)

and

|\vec{r}+\vec{p}|=4

\Rightarrow |\vec{r}|^{2}+\vec{r}\cdot \vec{p}+\vec{p}\cdot \vec{r}+|\vec{p}|^{2}=16

.....(vii)

Adding Eqs. (v), (vi) and (vii), we get

2|\vec{p}|^{2}+2|\vec{q}|^{2}+2|\vec{r}|^{2}+2(\vec{p}\cdot \vec{q}+\vec{q}\cdot \vec{r}+\vec{r}\cdot \vec{p})

=100

\Rightarrow |\vec{p}|^{2}+|\vec{q}|^{2}+|\vec{r}|^{2}= \frac{100}{2}=50

....(viii)

[using Eq.(iv)]

Now,

(\vec{p}+\vec{q}+\vec{r})^{2}

=|\vec{p}|^{2}+|\vec{q}|^{2}+|\vec{r}|^{2}+2(\vec{p}\cdot \vec{q}+\vec{q}\cdot \vec{r}+\vec{r}\cdot \vec{p})

=50

[using Eqs. (iv) and (viii)]

\Rightarrow |\vec{p}+\vec{q}+\vec{r}|=5\sqrt{2}

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